Is $x^4 -4x^3+6x^2-4x-8$ irreducible over $\mathbb{Q}$?

Question

If I let $z = x-1$ then $f(z) = z^4 - 9$. If I apply Eisenstein criterion with $p = 7$ I get that $f(z)$ is irreducible, but I can express $f(z) = (z^2-3)(z^2+3)$ so that means it is reducible. Am I applying Eisenstein criterion wrong?

Answer 1

Eisenstein's criterion demands

• $p\mid a_i$ for $i < n$
• $p^2 \nmid a_0$
• $p \nmid a_n$

Specifically, $7\nmid -9 = a_0$ is violated in the substituted case. In fact the condition for $a_0$ isn't satisfied for any $p$ because $p\mid -9 \Rightarrow p = 3 \Rightarrow p^2 = 9 \mid -9$.

A similar discourse leads to EC being unapplicable for any prime to the original polynomial.

Answer 2

Yes\ I don't know what is your version of Eisenstein criterion. Actually the Eisenstein criterion with $p=7$, would suppose $-4, 6, -4 $ and $-8$ are divisible by $7$ and $-8$ is not divisible by $7^2$. Only the last assertion is true.

Answer 3

You are not applying Eisenstein's criterion correctly. $7$ does not divide $-9$.

To apply Eisenstein's criterion (which is not applicable here, because the polynomial is reducible), you must find a prime $p$ which divides $-9$, such that $p^2$ does not divide $-9$. Moreover, $p$ should not divide the coefficient of $x^4$ which is $1$ here.

An instance of E.C is: $f(x) = x^2 - 14$. $7$ divides $14$ but its square does not, and $7$ does not divide $1$. So $f(x)$ is irreducible (in $\mathbb Q[x]$).